#### Answer

(a) $L = 30~m$
(b) The new current is $1.0~A$

#### Work Step by Step

(a) We can find the resistance:
$R = \frac{V}{I}$
$R = \frac{1.5~V}{0.50~A}$
$R = 3.0~\Omega$
We can find the required length:
$R = \frac{\rho~L}{A}$
$L = \frac{R~A}{\rho}$
$L = \frac{R~\pi~r^2}{\rho}$
$L = \frac{(3.0~\Omega)~(\pi)~(0.30\times 10^{-3}~m)^2}{2.8\times 10^{-8}~\Omega~m}$
$L = 30~m$
(b) If the wire is half this length, then the resistance $R$ is also half the original value. Then, according to $V = IR$, the current must double since the voltage remains the same. The new current is $1.0~A$